Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;

class Main{
  public static void main(String args[]){
    Scanner in = new Scanner(System.in);
    int n = in.nextInt();
    double p[] = new double[n];
    for(int i=0;i<n;i++) p[i] = in.nextDouble();
    double dp[][]=  new double[n][n+2];
    dp[0][1] = 1-p[0];
    dp[0][2] = p[0];
    for(int i=1;i<dp.length;i++)
      for(int j=1;j<dp[i].length;j++)
        dp[i][j] = dp[i-1][j-1]*p[i] + dp[i-1][j]*(1-p[i]);
    double ans = 0;
    for(int i=(int)Math.ceil(n/2) + 2;i<=n+1;i++)
      ans += dp[n-1][i];
 
    System.out.println(ans);
  }

  public static void print(double a[][]){
    for(int i=0;i<a.length;i++)
      System.out.println(Arrays.toString(a[i]));
  }
}